A pendulum with a bob of mass 5 kg is held stationary at a height h metres above the ground - NSC Physical Sciences - Question 5 - 2016 - Paper 1

Using conservation of energy, the potential energy (PE) lost by the bob when it swings down to point A is equal to the kinetic energy gained by the block:

$mgh = \frac{1}{2} m v^2$

Substituting for h:

$h = \frac{(\frac{1}{2} (5)(4.95)^2)}{(5)(9.81)} = 0.67 \text{ m}$

The work-energy theorem states that the net work done on an object is equal to the change in the object's kinetic energy.

The work done by the frictional force (W_f) can be calculated using:

$W_f = KE_{initial} - KE_{final}$

Where:

• $KE_{initial} = \frac{1}{2} m v^2$ at point B, where $v = 4.95 \, m/s$ and $m = 2 \, kg$.
• $KE_{final} = \frac{1}{2} m (2)^2$ at point C.

Calculating these values gives:

• $W_f = (\frac{1}{2}(2)(4.95)^2) - (\frac{1}{2}(2)(2)^2) = 10.7 \text{ J}$